The limit of Balmer series is 8646 $\text{Å}$ The wavelength of the first member of this series will be

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{{\mathrm{n}}_1}^2}-\frac{1}{{{\mathrm{n}}_2}^2}\right] \\ \mathrm{For} \mathrm{the} \mathrm{limit} \mathrm{of} \mathrm{Balmer} \mathrm{series}, \mathrm{we} \mathrm{have} \\ \frac{1}{{\mathrm{\lambda }}_{lim}}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{\mathrm{\infty }}\right]=\frac{\mathrm{R}}{4} \\ \mathrm{and} \frac{1}{{\mathrm{\lambda }}_{\mathrm{first} }}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5\mathrm{R}}{36} \\ {\mathrm{\lambda }}_{\mathrm{first} }={\mathrm{\lambda }}_{lim}\left(\frac{36}{5\times 4}\right)=3646\times \frac{9}{5}=6563\mathrm{Å} \end{gathered}$$